Difference between revisions of "2017 USAJMO Problems/Problem 1"
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Prove that there are infinitely many distinct pairs <math>(a,b)</math> of relatively prime integers <math>a>1</math> and <math>b>1</math> such that <math>a^b+b^a</math> is divisible by <math>a+b</math>. | Prove that there are infinitely many distinct pairs <math>(a,b)</math> of relatively prime integers <math>a>1</math> and <math>b>1</math> such that <math>a^b+b^a</math> is divisible by <math>a+b</math>. | ||
− | == | + | ==Solution 1== |
Let <math>a = 2^n - 1</math> and <math>b = 2^n + 1</math>. We see that <math>a</math> and <math>b</math> are relatively prime (they are consecutive positive odd integers). | Let <math>a = 2^n - 1</math> and <math>b = 2^n + 1</math>. We see that <math>a</math> and <math>b</math> are relatively prime (they are consecutive positive odd integers). | ||
Revision as of 19:26, 19 April 2017
Problem
Prove that there are infinitely many distinct pairs of relatively prime integers and such that is divisible by .
Solution 1
Let and . We see that and are relatively prime (they are consecutive positive odd integers).
Lemma: .
Since every number has a unique modular inverse, the lemma is equivalent to proving that . Expanding, we have the result.
Substituting for and , we have where we use our lemma and the Euler totient theorem: when and are relatively prime.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2017 USAJMO (Problems • Resources) | ||
First Problem | Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |